Solve for constants a and b given the limit of the function

[3ephemeralwnd]

Homework Statement

Determine all the numbers a and b for which the limit of the function as x approaches 2 equals to 4

sorry, I am kinda new at this, not sure how to type it out

 

[fzero]

Does writing

$$\sqrt{ax +b} -3 = \sqrt{a(x-2) + b+2a} -3$$

give you any ideas?

 

[eibon]

try multiply by its Conjugate that will help you

 

[3ephemeralwnd]

Oh, thank you
Ive gotten to a point where (x-2) cancels out, and i can sub in x=2 without the denominator coming out as undefined..
But I'm still not sure where to go from here

 

[fzero]

Oh, thank you
Ive gotten to a point where (x-2) cancels out, and i can sub in x=2 without the denominator coming out as undefined..
But I'm still not sure where to go from here

Can you explain a bit more of your work? If you're using my hint then you expanded the square root in powers of (x-2) and there's a divergent term, a term that's independent of x, and then terms that vanish as x -> 2. The coefficients of the first 2 terms can be thought of as a pair of equations to solve for a and b once you demand that the divergent part vanishes and that the finite term gives the correct limit.

 

[nddancer]

Does writing

$$\sqrt{ax +b} -3 = \sqrt{a(x-2) + b+2a} -3$$

give you any ideas?

Hi fzero, how did you get that second part?
Thanks!

 

[nddancer]

I'm not sure how sqr.rt.((x-2) + b + 2a) shows me how to get there from the original numerator. What did you multiply, or how did you substitute?

 

[fzero]

I'm not sure how sqr.rt.((x-2) + b + 2a) shows me how to get there from the original numerator. What did you multiply, or how did you substitute?

All I did was add and subtract 2 a under the square root so that I could express the numerator in terms of the x-2 that appeared in the denominator. However, if you're not familiar with series expansions, the next step would be far from obvious.

 

[nddancer]

I figured out how you rewrote the numerator, and I was able to eliminate the (x-2) in the numerator and denominator. I am now left with (3a+b-3)/sqr.rt.(b+2a+3) = 4. Do you have any hints as to what comes after that?

 

[fzero]

I figured out how you rewrote the numerator, and I was able to eliminate the (x-2) in the numerator and denominator. I am now left with (3a+b-3)/sqr.rt.(b+2a+3) = 4. Do you have any hints as to what comes after that?

There's also a term proportional to 1/(x-2). This gives a 2nd equation to use to solve for a,b.